Answer:
Step-by-step explanation:
xy = k
where k is the constant of variation.
We can also express the relationship between x and y as:
y =
where k is the constant of variation.
Since k is constant, we can find k given any point by multiplying the x-coordinate by the y-coordinate. For example, if y varies inversely as x, and x = 5 when y = 2, then the constant of variation is k = xy = 5(2) = 10. Thus, the equation describing this inverse variation is xy = 10 or y = .
Example 1: If y varies inversely as x, and y = 6 when x = , write an equation describing this inverse variation.
k = (6) = 8
xy = 8 or y =
Example 2: If y varies inversely as x, and the constant of variation is k = , what is y when x = 10?
xy =
10y =
y = × = × =
k is constant. Thus, given any two points (x1, y1) and (x2, y2) which satisfy the inverse variation, x1y1 = k and x2y2 = k. Consequently, x1y1 = x2y2 for any two points that satisfy the inverse variation.
Example 3: If y varies inversely as x, and y = 10 when x = 6, then what is y when x = 15?
x1y1 = x2y2
6(10) = 15y
60 = 15y
y = 4
Thus, when x = 6, y = 4.
2nd answer choice
constant of variation is xy. XY=23. If X=7 then Y=23/7.
Answer:
The data we have is:
The acceleration is 3.2 m/s^2 for 14 seconds
Initial velocity = 5.1 m/s
initial position = 0m
Then:
A(t) = 3.2m/s^2
To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.
V(t) = (3.2m/s^2)*t + 5.1 m/s
To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)
P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t
Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.
P(14s) = (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m
So the final position is 385 meters ahead the initial position.
Answer:
D. Subtract 4 from both sides
Explanation:
I just learned about quadratic equations in school. The first step of solving quadratic equations is to make sure it is in standard form and equal to 0.
Standard form:
ax^2 + bx + c = 0
